Question

find the 4th Taylor Polynomial of f(x)=ln(1+x)

Answer 1

is in 3rd derivative of it?

Answer 2

f = ln(1+x)

f' = (1+x)^(-1)

f'' = -(1+x)^(-2)

f''' = 2(1+x)^(-3)

Answer 3

Taylor = f/0! + f'/1! x + f''/2! x^2 + f'''/3! x^3 + ...

now i tend to get these mixed in with the Mclaurin tho so what am i missing?

Answer 4

if we had an x value to calibrate the derivatives then we could wrap this around a stable value

Answer 5

thats it, the x value is the "a" in the textbooks ...

\[Taylor = \frac{f(a)}{0!}+\frac{f'(a)}{1!}(x-a)^+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...\]

Answer 6

the Mclaurin is the same but with a=0

Answer 7

thank you, my problem was when I found the 4th derivative, I got 0

Answer 8

4th derivative or 4th poly in the series? the terminology has me at a loss.

what are you using for the "a" part? or is that left generic ...

Answer 9

you have to find the 4th deriv. to find the 4th polynomial. They didn't give me an a so I assumed a=0, most of the problems are that way in the section we are doing

Answer 10

f = ln(1+0)

f' = (1+0)^(-1)

f'' = -(1+0)^(-2)

f''' = 2(1+0)^(-3)

f'''' = -6(1+0)^-4

all the negative exponents tell us that the denominator part goes to 1 in the f^n stuff

ln(1+x) = 0 + x - 1/2 x^2 + 1/3 x^3 - 1/4 x^4

is what i get with 4 derivatives and simplified